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Lecture 1 f:VARIANTS OF SIMPLEX METHOD
Jeff Chak-Fu WONG
Department of Mathematics
Chinese University of Hong Kong
MAT581 SSMathematics for Logistics
Produced by Jeff Chak-Fu WONG 1
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Example 1 Use two phase simplex method to
Miximise z = 2x 1 x 2
subject tox 1 + x2 2
x 1 + x2 4
x 1 , x 2 0
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Solution:
Convert the LPP into maximisation by using
min z = max ( z),
the LPP becomes
Maximise ( z) = 2 x 1 + x2
subject to
x 1 + x2 2
x 1 + x2 4x 1 , x 2 0
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By introducing slack variable s 2 , surplus variable s 1 , articial variableA1 , the standard form of LPP is
max ( z) = 2 x 1 + x2
subject to
x 1 + x2 s1 + 0 s 2 + A1 = 2
x 1 + x2 + 0 s 1 + s2 = 4
x 1 , x 2 , s 1 , s 2 , A 1 0
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The initial basis matrix is:
B = sA 1 s2 = I 2 .
An obvious initial basis feasible solution is s2 = 4 , A 1 = 2 , i.e.,
X B = B 1
b = I 2 b = [2 4]T
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y 1 = B 1
a 1 = I 2 [1 1]T = [1 1]T
y 2 = B 1
a 2 = I 2 [1 1]T = [1 1]T
y 1 = B 1
s 1 = I 2 [ 1 0]T = [ 1 0]T
y 2 = B 1
s 2 = I 2 [0 1]T = [0 1]T
y A 1 = B 1
s A 1 = I 2 [1 0]T = [1 0]T
z1 c1 = C B x 1 c1 = [ 1 0]1 1]T 0 = 1
z2 c2 = C B x 2 c2 = [ 1 0][1 1]T 0 = 1
z3 c3 = C B s 1 c3 = [ 1 0][ 1 0]T
0 = 1z4 c4 = C B s 2 c4 = [ 1 0][0 1]T 0 = 0
z5 c5 = C B y A 1 c5 = [ 1 0][0 1]T ( 1) = 0
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Phase I The objective function of the auxiliary LPP ismax ( z ) = A1 .
Starting Tableau:Tableau 6.1.1
cj : (0 0 0 0 -1)
y 1 y 2 y 3 y 4 y A 1
C B X B x1
x2
s1
s2
A1
Min. Ratio-1 A1 2 1 1 -1 0 1 2/1=2
0 s2 4 1 1 0 1 0 4/1=4
zj cj -1 -1 1 0 0
From Tableau 6.1.1 we observe that
the non-basic variable x1 enters into the basis;
the basic variable A1 leaves the basis.
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Starting Tableau:Tableau 6.1.1
cj : (0 0 0 0 -1)
C B X B x1 x2 s1 s2 A1 Min. Ratio-1 A1 2 1 1 -1 0 1 2/1=2
0 s2 4 1 1 0 1 0 4/1=4 R2 R 1
zj cj
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Starting Tableau:Tableau 6.1.1
cj : (0 0 0 0 -1)
C B X B x1 x2 s1 s2 A1
0 x1 2 1 1 -1 0 1
0 s2 2 0 0 0 1 1
zj cj
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The new basis matrix is:
B = sA 1 s2 = I 2 B = a 1 s2 .
Find the inverse of B , i.e., B 1 , we have
B I 2 I 2 B 1 .
That is,
1 0 1 0
1 1 0 1 1 0 1 0
0 1 1 1 .
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y 1 = B 1
a 1 = [1 0]T
y 2 = B 1
a 2 = [1 0]T
y 3 = B 1 s 1 = [ 1 1]T
y 4 = B 1
s 2 = [0 1]T
z1 c1 = C B y 1 c1 = [0 0][1 0]T 0 = 0
z2 c2 = C B y 2 c2 = [0 0][1 0]T 0 = 0
z3 c3 = C B y 3 c3 = [0 0][ 1 1]T
0 = 0z4 c4 = C B y 4 c4 = [0 0][0 1]T 0 = 0
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First iteration:
Tableau 6.1.2cj : (0 0 0 0)
C B X B x1 x2 s1 s2
0 x1 2 1 1 -1 0
0 s2 2 0 0 1 1
zj cj z (X B ) = 0 0 0 0 0
Since all zj cj 0 and no articial variable appears in theoptimum basis, the current basic feasible solution is optimal to theauxiliary LPP and we proceed to Phase II .
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Phase II Assign the actual costs to the original variables. The newobjective function then becomes
Maximise ( z) = 2 x1
+ x2
+ 0 s1
+ 0 s2
The initial basic feasible solution for this phase is the one obtained atthe end of Phase I .
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The initial basis matrix is:
B = B = a 1 s2 .
An obvious initial basis feasible solution is x1 = 2 , s 2 = 2 , i.e.,
X B = B 1
b = [2 2]T
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y 1 = [1 0]T
y 2 = [1 0]T
y 3 = [ 1 1]T
y 4 = [0 1]T
z1 c1 = C B y 1 c1 = [2 0][1 0]T ( 2) = 0
z2 c2 = C B y 2 c2 = [2 0][1 0]T ( 1) = 1
z3 c3 = C B y 3 c3 = [2 0][ 1 1]T
0 = 2z4 c4 = C B y 4 c4 = [2 0][0 1]T 0 = 0
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The iterative simplex tableaus for this phase is:
Initial iteration:Tableau 6.1.3cj : (2 1 0 0)
y 1 y 2 y 3 y 4
C B X B x1 x2 s1 s2 Min. Ratio2 x1 2 1 1 -1 0
0 s2 2 0 0 1 1 2/1=2
z
j cj z(X B ) = 4 0 1 -2 0From Tableau 6.1.3, we observe that s1 enters into the basis and s 2leaves the basis.
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The iterative simplex tableaus for this phase is:
Initial iteration:Tableau 6.1.3cj : (2 1 0 0)
y 1 y 2 y 3 y 4
C B X B x1 x2 s1 s2 Min. Ratio2 x1 2 1 1 -1 0
0 s2 2 0 0 1 1 2/1=2
z
j cj z(X B ) = 4 0 1 -2 0From Tableau 6.1.3, we observe that s1 enters into the basis and s 2leaves the basis.
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The iterative simplex tableaus for this phase is:
Initial iteration:Tableau 6.1.3cj : (2 1 0 0)
y 1 y 2 y 3 y 4
C B X B x1 x2 s1 s2 Min. Ratio2 x1 2 1 1 -1 0
0 s2 2 0 0 1 1 2/1=2
z
j cj z(X B ) = 4 0 1 -2 0From Tableau 6.1.3, we observe that s1 enters into the basis and s 2leaves the basis.
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Tableau 6.1.3cj : (2 1 0 0)
C B X B x1 x2 s1 s2 Min. Ratio
2 x1 2 1 1 -1 0 R1 + R2
0 s2 2 0 0 1 1 2/1=2
zj cj z(X B ) = 4 0 1 -2 0 R3 + 2 R 2
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Tableau 6.1.3cj : (2 1 0 0)
C B X B x1 x2 s1 s2 Min. Ratio
2 x1 4 1 1 0 10 s1 2 0 0 1 1
zj cj z(X B ) = 8 0 1 0 2
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First iteration:Tableau 6.1.4
cj : (2 1 0 0)
C B X B x1 x2 s1 s2
2 x1 4 1 1 0 1
0 s1 2 0 0 1 1
z
j cj z(X B ) = 8 0 1 0 2
Since all zj cj 0, the current basic feasible solution is optimal.
The optimal solution is
x 1 = 4 , x 2 = 0 with Maximum of ( z) = 8
That is, x1 = 4 , x 2 = 0 with minimum of z = 8
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VARIANTS OF SIMPLEX METHOD
ARIANTS OF SIMPLEX METHOD 23
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In this section we discuss some complications and variations which
are very often encountered during simplex procedure.The following are some of them:
1. Degeneracy and cycling
2. Unbounded solution
3. Multiple solution or alternative optimum solution
4. No feasible solution
5. Unrestricted variables.
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1. DEGENERACY (TIE FOR MINIMUM RATIO)
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But sometimes this ratio may not be unique
(i.e. more than one variable is eligible to leave the basis)
or
at the very rst iteration, the value of one or more basic variablesin X B column becomes zero ,
this causes the problem of degeneracy.
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If the minimum ratio is zero, for two or more basic variables,degeneracy may result and the simplex routine to cycleindenitely.
That is, the solution which we have obtained in one iterationmay repeat again after few iterations and therefore, nooptimum solution may be obtained.
This concept is known as cycling or circling .
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Step 1: First nd out the rows for which the minimum non-negativeratio is the same (tie), for example, suppose there is a tie betweenrst and second row.
Step 2: Now, rearrange the columns of the usual simplex table sothat the columns forming the original unit matrix come rst.
Step 3: Find the minimum of the ratio
elements of rst column of unit matrixcorresponding elements of key column
Only for the rows for which the minimum ratio was not unique(i.e. for tied rows, in our example for rst and second rows).
(i) If this minimum is attained for second row (say) then this row willdetermine the key element by intersecting with the key column.
(ii) If this minimum is also not unique, then go to the next step.
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Step 4: Compute the minimum of the ratioelements of second column of unit matrix
corresponding elements of key column
only for the rows for which minimum ratio is not unique in Step 3
.If the minimum is also not unique, then go to next step.
Step 5: Next compute the minimum of the ratio
elements of third column of unit matrixcorresponding elements of key column
If this minimum is still not unique, then go on repeating the
above procedure till the unique minimum ratio is obtained toresolve the degeneracy.
After the resolution of this tie, simplex method is applied toobtain the optimum solution.
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Remark 1 If we have a tie for an articial variable and some other
variable we can choose the articial variable to leave the basis.
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Example 2 Solve the LPP
Maximise z = 3 x 1 + 9 x 2
subject to
x 1 + 4 x 2 8
x 1 + 2 x 2 4
x 1 , x 2 0
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Solution:
Introduce the slack variables s1
, s2
0, the LPP becomesMaximise z = 3 x 1 + 9 x 2 + 0 s 1 + 0 s 2
subject to
x 1 + 4 x 2 + s1 = 8x 1 + 2 x 2 + + s2 = 4
x 1 , x 2 , s 1 , s 2 0
An obvious initial basic feasible solution is s1 = 8 , s 2 = 4 .
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Tableau 6.2.1cj : (3 9 0 0)
C B X B x1 x2 s1 s2 Min. Ratio
0 s1
8 1 4 1 00 s2 4 1 2 0 1
zj cj z(X B ) -3 -9 0 0
From Tableau 6.2.1, we observe that the non-basic variable x2enters into the basis.
Since the minimum ratio is 2 for both the slack variables s1 ands 2 , there is a tie for the variable to leave the basis.
This is an indication for the existence of degeneracy in the givenLPP.
Rearrange the columns of the simplex table so that the initial
identity matrix appears rst.
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Tableau 6.2.1cj : (3 9 0 0)
C B X B x1 x2 s1 s2 Min. Ratio
0 s1
8 1 4 1 00 s2 4 1 2 0 1
zj cj z(X B ) -3 -9 0 0
From Tableau 6.2.1, we observe that the non-basic variable x2enters into the basis.
Since the minimum ratio is 2 for both the slack variables s1 ands 2 , there is a tie for the variable to leave the basis.
This is an indication for the existence of degeneracy in the givenLPP.
Rearrange the columns of the simplex table so that the initialidentity matrix appears rst.
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Tableau 6.2.1cj : (3 9 0 0)
C B X B x1 x2 s1 s2 Min. Ratio
0 s1
8 1 4 1 0 8/4= 20 s2 4 1 2 0 1 4/2= 2
zj cj z(X B ) -3 -9 0 0
From Tableau 6.2.1, we observe that the non-basic variable x2enters into the basis.
Since the minimum ratio is 2 for both the slack variables s1 ands 2 , there is a tie for the variable to leave the basis.
This is an indication for the existence of degeneracy in the givenLPP.
Rearrange the columns of the simplex table so that the initialidentity matrix appears rst.
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Tableau 6.2.1cj : (3 9 0 0)
C B X B x1 x2 s1 s2 Min. Ratio
0 s1
8 1 4 1 0 8/4= 20 s2 4 1 2 0 1 4/2= 2
zj cj z(X B ) -3 -9 0 0
From Tableau 6.2.1, we observe that the non-basic variable x2enters into the basis.
Since the minimum ratio is 2 for both the slack variables s1 ands 2 , there is a tie for the variable to leave the basis.
This is an indication for the existence of degeneracy in the givenLPP.
Rearrange the columns of the simplex table so that the initialidentity matrix appears rst.
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Now, using the Step 3 of the procedure for resolving degeneracy,we nd
min elements of rst column
corresponding elements of key column
= min 14
, 02
= 0 .
which occurs for the second row. Hence, s2 must leave the basisand the pivot element is 2.
Tableau 6.2.2cj : (0 0 3 9)
C B X B s1 s2 x1 x2 Min. Ratio0 s1 8 1 0 1 4 1/4
0 s2 4 0 1 1 2 0/2=0
zj cj z(X B ) = 0 0 0 -3 -9
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Now, using the Step 3 of the procedure for resolving degeneracy,we nd
min elements of rst column
corresponding elements of key column
= min 14
, 02
= 0 .
which occurs for the second row. Hence, s2 must leave the basisand the pivot element is 2.
Tableau 6.2.2cj : (0 0 3 9)
C B X B s1 s2 x1 x2 Min. Ratio0 s1 8 1 0 1 4 1/4
0 s2 4 0 1 1 2 0/2=0
zj cj z(X B ) = 0 0 0 -3 -9
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Tableau 6.2.2cj : (0 0 3 9)
C B X B s1 s2 x1 x2 Min. Ratio
0 s1 8 1 0 1 4 1/4
0 s2 4 0 1 1 2 0/2=0 12 R 2
zj cj z(X B ) = 0 0 0 -3 -9
Tableau 6.2.2cj : (0 0 3 9)
C B X B s1 s2 x1 x2 Min. Ratio
0 s1 0 1 -2 -1 0 1/4 R1 4 R 2
0 s2 2 0 1/2 1/2 1 0/2=0
zj cj z(X B ) = 18 0 9/2 3/2 0 R3 + 9 R 2
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First iteration:Tableau 6.2.3
cj : (0 0 3 9)
C B X B s1 s2 x1 x2
0 s1 0 1 -2 -1 0
9 x2 2 0 1/2 1/2 1
zj cj z(X B ) = 18 0 9/2 3/2 0
Since all zj cj 0, an optimum solution has been reached.
Hence, the optimum basic feasible solution is
x 1 = 0 , x 2 = 2 with maximum of z = 18
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2. Unbounded solutions: The case of unbounded solutions occur when the feasible
region is unbounded such that the value of the objectivefunction can be increased indenitely.
It is not necessary, however, that an unbounded feasible regionwill give rise to an unbounded value of the objective function.
The following example will illustrate these points.
. Unbounded solutions: 44
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UNBOUNDED SOLUTION SPACE BUT BOUNDED OPTIMAL SOLUTION
Example 3 Solve the following LPP
Maximise z = 3 x 1 x 2
subject to
x 1 x 2 10
x 1 20
x 1 , x 2 0
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Solution:
By introducing the slack variables s1 , s 2 0, the LPP becomes
Maximise z = 3 x 1 x 2 + 0 s 1 + 0 s 2
subject to
x 1 x2 + s1 = 10
x 1 + s2 = 20
x 1 , x 2 , s 1 , s 2 0
An obvious initial basic feasible solution is s1 = 10 , s 2 = 20 .
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Tableau 6.3.3c j : (3 -1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
3 x 1 10 1 -1 1 00 s 2 10 0 1 -1 1
z j c j z(X B ) = 30 0 -2 3 0
Second Iteration
Tableau 6.3.4c j : (3 -1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
3 x 1 10 1 -1 1 0
0 s 2 10 0 1 -1 1 10/1 = 10
z j c j z(X B ) = 30 0 -2 3 0
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Tableau 6.3.5 c j : (3 -1 0 0)y 1 y 2 y s 1 y s 2
C B X B x 1 x 2 s 1 s 2 Min. Ratio
3 x 1 10 1 -1 1 0 R 1 + R 20 s 2 10 0 1 -1 1 10/1 = 10
z j c j z(X B ) = 30 0 -2 3 0 R 3 + 2 R 2
Tableau 6.3.6c j : (3 -1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
3 x 1 20 1 0 0 1
-1 x 2 10 0 1 -1 1
zj
cj
z(X
B ) = 50 0 0 1 2
The optimum solution is:
x 1 = 20 , x 2 = 10 with maximum of z = 50
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The graphical solution of the LPP is given in Figure 1.
0
x = 20
x - x = 10
B(20,10)
A(10,0)
15105 20
5
10
x1
1
1 2
x2
15
(a)
0
x = 20
x - x = 10
B(20,10)
A(10,0)
15105 20
5
10
x1
1
1 2
x2
15
(b)
0
x = 20
x - x = 10
B(20,10)
A(10,0)
15105 20
5
10
x1
1
1 2
x2
15
(c)
Figure 1:
We observe that the solution space is unbounded. But the optimalsolution occurs at the vertex (20, 10) .
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BOUNDED SOLUTION SPACE AND UNBOUNDED O PTIMAL SOLUTION
Example 4 Solve the following LPP
Maximise z = 2 x 1 + x2
subject tox 1 x 2 10
2x 1 x 2 40
x 1 0, x 2 0
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Solution:
Introducing the slack variables s 1 , s 2 0 the LPP in standard form is
Maximise z = 2 x 1 + x2 + 0 s 1 + 0 s 2
subject to
x 1 x2 + s1 = 10
2x 1 x2 + s2 = 40
x 1 , x 2 , s 1 , s 2 0
An obvious IBFS is s1 = 10 , s 2 = 40 .
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Starting Table:Tableau 6.4.1
c j : (2 1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
0 s 1 10 1 -1 1 0
0 s 2 40 2 -1 0 1
z j c j z(X B ) = 0 -2 -1 0 0
Tableau 6.4.2c j : (2 1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
0 s 1 10 1 -1 1 0 10
1 = 100 s 2 40 2 -1 0 1 402 = 20
z j c j z(X B ) = 0 -2 -1 0 0
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Tableau 6.4.3c j : (2 1 0 0)
y 1 y 2 y s 1 y s 2
C B X B x 1 x 2 s 1 s 2 Min. Ratio0 s 1 10 1 -1 1 0
0 s 2 40 2 -1 0 1 R 2 2 R 1
z j c j z(X B ) = 0 -2 -1 0 0 R 3 + 2 R 1
Tableau 6.4.4c j : (2 1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
2 x 1 10 1 -1 1 0
0 s 2 20 0 1 -2 1
z j c j z(X B ) = 20 0 -3 2 0
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Second iteration:
Table 6.4.8c j : (2 1 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
2 x 1 30 1 0 -1 1
1 x 2 20 0 1 -2 1
z j c j z(X B ) = 80 0 0 -4 3
From Table 6.4.8 we see that s 1 column is the pivotal column,but there is no positive element in that column.
Hence, there exists an unbounded solution to the given LPP. If we solve the LPP by graphical method we can see that the
feasible region is unbounded. See Figure 2.
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0x1
x2
5 10 20
10
20
30
40
10
20
P(30,20)
x - x = 101 2
2 x - x = 401 2
30
Figure 2:
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3. M ULTIPLE SOLUTIONS OR ALTERNATE O PTIMAL SOLUTIONS While solving the LPP by simplex method, in the optimum simplex
table, if the net evaluation zj cj = 0 for all non-basic variables,then the problem is said to have a unique optimal solution.
On the other hand, if the net evaluation zj cj = 0 for at leastone non-basic variable, then the problem is said to have analternative or innite number of solutions.
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In a graphical method if the optimal solution occurs at a vertexof the solution space, then the problem said to havea unique optimal solution.
If the optimum solution occurs on an edge of the solutionspace, then the problem is said to have an alternative or innitenumber of solutions .
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MULTIPLE SOLUTIONS OR ALTERNATE O PTIMAL SOLUTIONS
Example 5 Solve the following LPP
Maximise z = x1 + 12
x 2
subject to2x 1 + x2 4
x 1 + 2 x 2 3
x 1 , x 2 0
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Solution:
By introducing the slack variables s1 , s 2 0, the LPP becomes
Maximise z = x1 + 12
x 2 + 0 s 1 + 0 s 2
subject to
2x 1 + x2 + s1 = 4
x 1 + 2 x 2 + s2 = 3
x 1 , x 2 , s 1 , s 2 0
An obvious initial basic feasible solution is s1 = 4 , s 2 = 3 .
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First Iteration
Tableau 6.5.1c j : (1
12 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
0 s 1 4 2 1 1 0
0 s 2 3 1 2 0 1
z j c j z(X B ) = 0 -1 12 0 0
Tableau 6.5.2c j : (1
12 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
0 s 1 4 2 1 1 0 42 = 2 12 R 1
0 s 2 3 1 2 0 1 31 = 4
z j c j z(X B ) = 0 -1 12 0 0
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Tableau 6.5.3
c j : (1 1
2 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
0 s 1 2 1 1/2 1/2 0 42 = 20 s 2 3 1 2 0 1 31 = 4 R 2 R 1
z j c j z(X B ) = 0 -1 12 0 0 R 3 + R 1
Tableau 6.5.4c j : (1 12 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
1 x 1 2 1 1/2 1/2 0
0 s 2 1 0 3/2 -1/2 1
z j c j z(X B ) = 2 0 0 1/2 0
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Tableau 6.5.5c j : (1
12 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
1 x 1 2 1 1/2 1/2 0 21 / 2 = 4
0 s 2 1 0 3/2 -1/2 1 13 / 2 = 2 / 3 2
3 R 2
z j c j z(X B ) = 2 0 0 1/2 0
Tableau 6.5.6
c j : (1 12 0 0)
y 1 y 2 y s 1 y s 2C B X B x 1 x 2 s 1 s 2 Min. Ratio
1 x 1 2 1 1/2 1/2 0 2
1 / 2 = 4 R 1 1
2 R 20 s 2 23 0 1 -1/3 2/3
13 / 2 = 2 / 3
z j c j z(X B ) = 2 0 0 1/2 0
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Tableau 6.5.7
c j : (1 12 0 0)
y 1 y 2 y s 1 y s 2
C B X B x 1 x 2 s 1 s 2 Min. Ratio1 x 1 53 1 0 2/3 -1/312 x 2
23 0 1 -1/3 2/3
z j c j z(X B ) = 2 0 0 1/2 0
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NO FEASIBLE SOLUTION OR NON - EXISTING FEASIBLE SOLUTION
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In an LPP, where there is no point in the solution space satisfyingall the constraints, then the problem is said to have no feasiblesolution .
In simplex method, if there exists at least one articial variable inthe basis at positive level, and even though optimalityconditions are satised which is the indication of non-feasiblesolution.
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C.f. Example 2 in Lecture Note 1d.
Maximise z = 3 x 1 + 2 x 2
subject to
2x 1 + x2 23x 1 + 4 x 2 12
x 1 , x 2 0.
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First iteration:Tableau 5.4.2
cj : (3 2 0 0 M )C B X B x1 x2 s1 s2 A1
2 x2 2 2 1 1 0 0
M A 1 4 -5 0 -4 -1 1
zj cj z(X B ) = 4M + 4 5M + 1 0 4M + 2 M 0
Here, the coefcient of M in each z j c j 0 , and an articial variableA 1 appears in the basis at non-zero level.
Thus, the given LPP does not possess any feasible solution.
We can say the LPP possess a pseudo-optimal solution.
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UNRESTRICTED VARIABLE
NRESTRICTED VARIABLE 72
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In an LPP, if any variable is unrestricted (it can have positivevalue or negative value or zero value) it can be expressed asthe difference between two non-negative variables.
The problem can be converted into an equivalent one involvingonly non-negative variables.
NRESTRICTED VARIABLE 73
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Example 6 Solve the LPP
Maximise z = 2 x 1 + 3 x 2
subject to
x 1 + 2 x 2 4x 1 + x2 6
x 1 + 3 x 2 9
and x1 , x 2 are unrestricted.
NRESTRICTED VARIABLE 74
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Tableau 6.6.1c j : (2 -2 3 -3 0 0 0)
y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 Min Ratio
0 s1
4 -1 1 2 -2 1 0 00 s 2 6 1 -1 1 -1 0 1 0
0 s 3 9 1 -1 3 -3 0 0 1
z j c j z(X B ) = 0 -2 2 -3 3 0 0 0
Tableau 6.6.2 c j : (2 -2 3 -3 0 0 0)y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3
C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 Min Ratio
0 s 1 4 -1 1 2 -2 1 0 0 4/2 = 2
0 s 2 6 1 -1 1 -1 0 1 0 6/1 = 60 s 3 9 1 -1 3 -3 0 0 1 9/3 = 3
z j c j z(X B ) = 0 -2 2 -3 3 0 0 0
NRESTRICTED VARIABLE 77
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Tableau 6.6.3c j : (2 -2 3 -3 0 0 0)
y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 Min Ratio
0 s 1 2 -1/2 1/2 1 -1 1/2 0 0 4/2 = 2 12 R 10 s 2 6 1 -1 1 -1 0 1 0 6/1 = 6
0 s 3 9 1 -1 3 -3 0 0 1 9/3 = 3
z j c j z(X B ) = 0 -2 2 -3 3 0 0 0
Tableau 6.6.4c j : (2 -2 3 -3 0 0 0)
y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 Min Ratio
3 x 2 2 -1/2 1/2 1 -1 1/2 0 0 4/2 = 2
0 s 2 4 3/2 -3/2 0 0 -1/2 1 0 R 2 R 10 s 3 3 5/2 -5/2 0 0 -3/2 0 1 R 3 3 R 1
z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0 R 4 + 3 R 1
NRESTRICTED VARIABLE 78
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Tableau 6.6.5c j : (2 -2 3 -3 0 0 0)
y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 Min Ratio
3 x 2 2 -1/2 1/2 1 -1 1/2 0 0
0 s 2 4 3/2 -3/2 0 0 -1/2 1 0
0 s 3 3 5/2 -5/2 0 0 -3/2 0 1
z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0
Tableau 6.6.6 c j : (2 -2 3 -3 0 0 0)y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3
C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 Min Ratio
3 x 2 2 -1/2 1/2 1 -1 1/2 0 0
0 s 2 4 3/2 -3/2 0 0 -1/2 1 0 4/(3/2) = 2.6667
0 s 3 3 5/2 -5/2 0 0 -3/2 0 1 3/(5/2) = 1.2 25 R 3
z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0
NRESTRICTED VARIABLE 79
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Tableau 6.6.7c j : (2 -2 3 -3 0 0 0)
y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 M. R.
3 x 2 2 -1/2 1/2 1 -1 1/2 0 0 R 1 + 1
2 R 30 s 2 4 3/2 -3/2 0 0 -1/2 1 0 R 2 32 R 3
0 s 3 6/5 1 -1 0 0 -3/5 0 2/5
z j c j z(X B ) = 6 -7/2 7/2 0 0 3/2 0 0 R 4 + 7
2 R 3
Tableau 6.6.8 c j : (2 -2 3 -3 0 0 0)y 1 y 2 y 3 y 4 y s 1 y s 2 y s 3
C B X B x
1 x
1 x
2 x
2 s 1 s 2 s 3 M. R.
3 x 2 13/5 0 0 1 -1 1/5 0 1/5
0 s 2 11/5 0 0 0 0 2/5 1 -3/5
2 x 1 6/5 1 -1 0 0 -3/5 0 2/5
z j c j z(X B ) = 51 / 5 0 0 0 0 -3/5 0 7/5
NRESTRICTED VARIABLE 80
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If we solve the LPP by ordinary simplex method, the optimumsolution is
x 1 = x 1 x 1 = 9 / 2 0 = 9 / 2
x 2 = x 2 x 2 = 3 / 2 0 = 3 / 2with maximum of z = 27 / 2
NRESTRICTED VARIABLE 83